Conservation of ang. momentum for paths reaching a rotation axis

[quantum52]

Hey everyone.
My question is the following: if we had the trajectory of a particle eventually reaching a point of a rotation axis [itex] \vec{u} [/itex] ((take that as being the z-axis for convenience) by an angle [itex] s [/itex] , would Noethers Theorem still give a conserved quantity?

More specifically (let me go through the calculations and details first)

1. Statement of Noether's Theorem

If a Lagrangian [itex] \mathcal{L}(\vec{q_i}, \dot{\vec{q_i}}, t) [/itex] admits a one-parameter group of diffeomorphisms [itex] h^s : \mathcal{M} \rightarrow \mathcal{M} [/itex] such that [itex] h^{(s=0)} (\vec{q_i})= \vec{q_i} [/itex], then there is a conserved quantity locally given by [tex] I = \sum_{i} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \left.\frac{d}{ds}(h^s(q_i))\right\vert_{s=0} [/tex]

2. Applying to Simple Lagrangian

Assume a potential-free Lagrangian [itex] \mathcal{L} = \frac{m}{2}( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 ) [/itex].
A suitable transformation can be given by [tex] h^s (x,y,z)= \begin{pmatrix}
cos(s) & -sin(s) & 0 \\
sin(s) & cos(s) & 0 \\
0 & 0 & 1
\end{pmatrix} [/tex]

Working out the conserved quantity, we get that the z-component of angular momentum [itex] L_z = m \dot{y}(t) x(t) - m \dot{x}(t) y(t) [/itex] is conserved for any path [itex] (x(t),y(t),z(t)) [/itex].

3. The problem:
If this trajectory would include any point on the rotation axis z, [itex] h^s(q_i) [/itex] would be 0 there and so by conservation, valued 0 all along the path.
However, we know that angular momentum is conserved.
So, in all rigour - is this inconsistency amendable or a sign of some bigger problem?

Any comments are very welcome.
David

 

[eljose79]

Hello David,

Thank you for your question. Noether's theorem is a powerful tool in physics and can provide insight into the underlying symmetries and conserved quantities of a system. In your scenario, if the particle is moving along a trajectory that eventually reaches the rotation axis, then the symmetry under rotation about that axis is broken and Noether's theorem would not apply in the same way.

In your calculations, you assumed a potential-free Lagrangian, which means there are no external forces acting on the particle. However, if the particle reaches the rotation axis, it would experience a centripetal force that would change its motion and break the symmetry. In this case, the conserved quantity would not be the z-component of angular momentum, but rather the total angular momentum of the system, which includes the angular momentum of the particle and the rotation axis.

In general, Noether's theorem is only applicable to systems with a continuous symmetry. In your scenario, the rotation symmetry is broken at the point where the particle reaches the rotation axis, so the theorem does not apply in the same way. However, the total angular momentum of the system is still conserved, as angular momentum is a conserved quantity in any system with rotational symmetry.

I hope this helps clarify the issue. If you have any further questions, please let me know.